Question

AB and CD are two chords of a circle,
such that AB = 6 cm, CD = 12 cm and
AB || CD. If the distance between AB and
CD is 3 cm, find the radius of the circle.​

Answer 1

Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm CD = 12 cm. Let the radius of the circle be r cm. Drwa OP ⊥ AB and OQ ⊥ CD. Since, AB ∥ CD and OP ⊥ AB, OQ ⊥ CD. Therefore points O, Q and P are collinear. Clearly, PQ = 3 cm.

Let OQ = x cm. Then, OP = (x + 3) cm

In right triangles OAP and OCQ, we have

OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2

⇒ r2 = (x + 3)2 + 32 and r2 = x2 + 62

[∵ AP = ½ AB = 3 cm and CQ = ½ CD = 6 cm

⇒ (x + 3)2 + 32 = x2 + 62 (on equating the value of r2)

⇒ x2 + 6x + 9 + 9 = x2 + 36

⇒ 6x = 18 ⇒ x = 3 cm

Putting the values of x in r2 = x2 + 62, we get

r2 = 32 + 62 = 45

⇒ r = √45 cm = 6.7 cm

Hence, the radius of the circle is 6.7 cm.